Expert: Steve Holleran - 3/20/2008

Question
Hi i'm not really sure how to solve this, can u help please?

The sum, S, of the first n positive integers, 1+2+3+4+...+n, is given by S=n/2(n+1). Find:

a) the value of 51+52+53+...+80
b) the number of positive integers needed to give a sum of 666


Answer
Hi Anna,

Okay, using your formula, the sum of the first 80 positive integers would be :

               s(80) = 80/2 * 81 = 40 * 81 = 3240

The sum of the first 50 would be :

               s(50) = 50/2 * 51 = 25 * 51 = 1275

So subtracting, you have 3240 - 1275 = 1965.



Then for part b, you have the equation

               666 = (n/2) * (n+1)

multiply by 2:   1332 = n * (n+1)

so                 0 = n^2 + n  - 1332   

and you can use the quadratic formula to solve and get

                    n = -1 +/- 73 / 2 = -74/2 or 72/2

The positive one is the one we want, so n = 72/2 = 36.


Hope this helps.

Steve